Separate variables: \(\frac{dy}{1-y^2} = \frac{dx}{x}\).

Express \(\frac{1}{1-y^2}\) as partial fractions: \(\frac{1}{1-y^2} = \frac{1}{2(1-y)} + \frac{1}{2(1+y)}\).

Integrate both sides: \(-\frac{1}{2} \ln(1-y) + \frac{1}{2} \ln(1+y) = \ln x + C\).

Combine logarithms: \(\frac{1}{2} \ln \left( \frac{1+y}{1-y} \right) = \ln x + C\).

Use initial condition \(x = 2, y = 0\) to find \(C\):

\(\frac{1}{2} \ln 1 = \ln 2 + C \Rightarrow C = -\ln 2\).

Substitute back: \(\frac{1}{2} \ln \left( \frac{1+y}{1-y} \right) = \ln x - \ln 2\).

Exponentiate to solve for \(y\):

\(\frac{1+y}{1-y} = \frac{x^2}{4}\).

Cross-multiply and solve for \(y\):

\(4(1+y) = x^2(1-y)\).

\(4 + 4y = x^2 - x^2y\).

\(4y + x^2y = x^2 - 4\).

\(y(4 + x^2) = x^2 - 4\).

\(y = \frac{x^2 - 4}{x^2 + 4}\).